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I have a mechanical system that requires a certain amount of electrical power. Now, when I calculated everything I did not include the internal resistance.

For example:

I calculated that the mechanical system will require exactly 100W of electrical power. Do I have to add more power to account for the losses due to the internal resistance of the power source?

I understand there are forms of resistance, of the circuit, of the mechanical system itself but what about the internal resistance? And it's affect on the power conversion?

If my system requires for example, 10A @ 10V, would I need to supply more power due to the internal resistance, a friend of mine pointed out that a lot of power is lost due to the internal resistance, and honestly I'm confused as to how.

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  • \$\begingroup\$ If your example power supply guarantees 10V at 10A then it takes its own internal resistance into account. (You'll find it consumes more than 100W, perhaps 1.0A or 1.1A at 110V). But you DO need to account for any resistance in your own wiring. \$\endgroup\$ – user_1818839 Jul 8 '14 at 9:07
  • \$\begingroup\$ Well, isn't that natural? A bit of energy is always lost due to resistance,etc... no system(built by us) can ever be 100% efficient, at best 90% efficient. \$\endgroup\$ – Pupil Jul 8 '14 at 9:32
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    \$\begingroup\$ Absolutely correct, but 90% isn't a fundamental upper limit; the Victorians could make 98% efficient transformers, but they used a lot of iron and copper. My point is just that if the output is guaranteed, accounting for losses is transferred to the input. \$\endgroup\$ – user_1818839 Jul 8 '14 at 10:39
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    \$\begingroup\$ Internal resistance is typically quite low and may only become a problem when your load resistance approaches this value. As Brian has said, the supply will consume more than 100W due to parasitic resitances, but if you are guarenteed 10A and 10V then the supply was designed with its output resistance in mind (i.e. higher voltage generated internally to account for 10A through output resistance so your circuit still sees 10V out). \$\endgroup\$ – sherrellbc Jul 8 '14 at 12:57
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Whatever the power supply might do internally is not relevant to you. If it says it will supply 10 A at 10 V, then it will supply 10 A at 10 V. Of course none of these supplies will be 100% efficient, so it will take more than 100 W in to produce that 100 W out. This extra input power has been taken into account in the spec that says what maximum current the supply will draw at certain line voltages. The fuse will also be sized accordingly.

Otherwise, a power supply is a "black box" that you shouldn't make assumptions about how it works internally. It specifies what goes in and what comes out, but other than that you don't know, and it's none of your business, what goes on inside to accomplish that.

As a separate topic, if your mechanical system requires exactly 100 W of power, then specifying a 100 W power supply is not leaving any room for anything to be a little degraded or for stuff to happen. This is bad engineering. What happens when the bearings get a little worn, there is a little more friction than you expect, etc, etc?

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  • \$\begingroup\$ I just used an example so that I can understand the role of internal resistance. \$\endgroup\$ – Pupil Jul 9 '14 at 0:51

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